a) [tex]\text{x}^2+3=4\text{x}[/tex]
Temos que:
[tex]\text{x}^2+3=4\text{x}[/tex]
[tex]\text{x}^2-4\text{x}+3=0[/tex]
[tex]\text{x}=\dfrac{-(-4)\pm\sqrt{(-4)^2-4\cdot1\cdot3}}{2\cdot1}=\dfrac{4\pm2}{2}[/tex]
Logo, as raízes são:
[tex]\text{x}'=\dfrac{4+2}{2}=3[/tex]
[tex]\text{x}''=\dfrac{4-2}{2}=1[/tex]
b) [tex]\text{x}\cdot(\text{x}+2)=3[/tex]
Temos que:
[tex]\text{x}^2+2\text{x}-3=0[/tex]
[tex]\text{x}=\dfrac{-2\pm\sqrt{2}^2-4\cdot1\cdot(-3)}}{2\cdot1}+\dfrac{-2\pm4}{2}[/tex]
Desta maneira, as raízes são:
[tex]\text{x}'=\dfrac{-2+4}{2}=1[/tex]
[tex]\text{x}"=\dfrac{-2-4}{2}=-3[/tex]
c) [tex]\text{x}(2\text{x}-1)+6=4\cdot(\text{x}+1)[/tex]
[tex]2\text{x}^2-\text{x}+6=4\text{x}+4[/tex]
[tex]2\text{x}^2-5\text{x}+2=0[/tex]
[tex]\text{x}=\dfrac{-(-5)\pm\sqrt{(-5)^2-4\cdot2\cdot2}}{2\cdot2}=\dfrac{5\pm3}{4}[/tex]
Logo, as raízes são:
[tex]\text{x}'=\dfrac{5+3}{4}=2[/tex]
[tex]\text{x}"=\dfrac{5-3}{4}=\dfrac{2}{4}=\dfrac{1}{2}[/tex]
